Question: Simplify; express your answer in exponential form. Assume $r\neq 0, k\neq 0$. $\dfrac{{(r^{-1})^{-5}}}{{(r^{2}k)^{2}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-1}}$ to the exponent ${-5}$ . Now ${-1 \times -5 = 5}$ , so ${(r^{-1})^{-5} = r^{5}}$ In the denominator, we can use the distributive property of exponents. ${(r^{2}k)^{2} = (r^{2})^{2}(k)^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-1})^{-5}}}{{(r^{2}k)^{2}}} = \dfrac{{r^{5}}}{{r^{4}k^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{5}}}{{r^{4}k^{2}}} = \dfrac{{r^{5}}}{{r^{4}}} \cdot \dfrac{{1}}{{k^{2}}} = r^{{5} - {4}} \cdot k^{- {2}} = rk^{-2}$.